An architect is allowed 15 square yards of floor space to add a small bedroom to a house. Because of the room's design in relationship to the existing structure, the width of the rectangular floor must be 7 yards less than two times the length. Find the length and width of the rectangular floor that the architect is permitted.

We are tasked with finding the length and width of the rectangular floor. The total floor area is given as 15 square yards, and the width of the rectangular floor is defined as 7 yards less than twice the length.

Let's define the length and width in terms of a variable:
- Let the length of the floor be \( L \) yards.
- The width of the floor is \( 2L - 7 \) yards (because it's 7 yards less than twice the length).

The area of a rectangle is given by the formula:
\[
\text{Area} = \text{Length} \times \text{Width}
\]
Substituting the given area and the expression for the width:
\[
15 = L \times (2L - 7)
\]
Now let's solve this equation:

### Step 1: Expand the equation
\[
15 = L(2L - 7) \Rightarrow 15 = 2L^2 - 7L
\]

### Step 2: Rearrange the equation into standard quadratic form
\[
2L^2 - 7L - 15 = 0
\]

### Step 3: Solve the quadratic equation
We can solve this using the quadratic formula, where \( a = 2 \), \( b = -7 \), and \( c = -15 \):
\[
L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Substitute the values of \( a \), \( b \), and \( c \):
\[
L = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(-15)}}{2(2)}
\]
\[
L = \frac{7 \pm \sqrt{49 + 120}}{4}
\]
\[
L = \frac{7 \pm \sqrt{169}}{4}
\]
\[
L = \frac{7 \pm 13}{4}
\]
This gives two possible solutions for \( L \):
\[
L = \frac{7 + 13}{4} = \frac{20}{4} = 5
\]
or
\[
L = \frac{7 - 13}{4} = \frac{-6}{4} = -1.5
\]
Since length must be positive, we discard \( L = -1.5 \). Therefore, the length is \( L = 5 \) yards.

### Step 4: Find the width
Substitute \( L = 5 \) into the expression for the width:
\[
\text{Width} = 2L - 7 = 2(5) - 7 = 10 - 7 = 3 \text{ yards}
\]

### Final Answer:
- The length of the rectangular floor is \( 5 \) yards.
- The width of the rectangular floor is \( 3 \) yards.

Would you like further clarification or details?

Here are 5 related questions for deeper understanding:
1. How do you derive the quadratic formula used to solve for \( L \)?
2. Why is the negative solution for \( L \) discarded in this context?
3. What other real-life applications might require solving similar quadratic equations?
4. How would the solution change if the total area allowed was different, say 20 square yards?
5. Can you confirm the solution by re-checking the area with the calculated length and width?

**Tip**: When solving quadratic equations, always check whether both solutions are valid within the problem’s context.

Learn how to find the dimensions of a rectangular floor given a total area of 15 square yards and a relationship between the length and width using quadratic equations. This problem involves solving a quadratic equation and is suitable for students in Grades 9-10.

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